Draw the flat torus

Question

I want to graph a curve on the flat torus, but not on the entire R^2 plane, but on the square [0,1]x[0,1].

For this, I have created a "do...while" with only the initial data of the rational slope. This is:

let m=p/q
let fx(u)=m(1-u) and fy(u)=(1-u)/m
x=0
DO
y=fx(x)
draw (x,0) -- (1,y);
x=fy(y)
draw (0,y) -- (x,1);
WHILE (x=1)

I apologize for the big mistakes in my code. I just wrote my idea since I don't know how to do it in LaTeX, using TikZ for example. How can I do it?

I already did it manually, but it is not the most optimal. Trying to put m=3/4 is supposed to output the following:

I don't want to code where I have to plot everything manually like the following:

\documentclass[border=1mm]{standalone}
\usepackage{tikz}
\begin{document}
\begin{tikzpicture}
\draw[red] (0,0) -- (1,3/4);
\draw[red] (0,3/4) -- (1/3,1);
\draw[red] (1/3,0) -- (1,1/2);
\draw[red] (0,1/2) -- (2/3,1);
\draw[red] (2/3,0) -- (1,1/4);
\draw[red] (0,1/4) -- (1,1);
\draw (0,0) -- (1,0) -- (1,1) -- (0,1) -- (0,0) -- (1,0);
\end{tikzpicture}
\end{document}

Answer 1

Here's an option. I defined a command \flattorus with three arguments:

1. (optional) The side of the square in the TikZ picture, in cm. I put 4 as the default value, but of course you can easily change that.
2. The number of columns to split the square into.
3. The number of rows to split the square into.

\documentclass{article}
\usepackage{tikz}
\newcounter{mx}
\newcounter{my}
\newlength{\squareside}
\newcommand*{\flattorus}[3][4]{%
    \setcounter{mx}{#2}
    \setcounter{my}{#3}
    \addtocounter{mx}{-1}
    \addtocounter{my}{-1}
    \setlength{\squareside}{#1 cm}

    \begin{tikzpicture}[x=\dimexpr\squareside/#2, y=\dimexpr\squareside/#3]
        \draw[thick] (0,0) rectangle (#2,#3);
        \foreach \x in {0, ..., \value{mx}}
            \foreach \y in {0, ..., \value{my}}{
                \draw (\x,\y) -- ++(0,1);
                \draw (\x,\y) -- ++(1,0);
                \draw[red, thick] (\x,\y) -- ++(1,1);
            };
        \node[below left] at (0,0) {0};
        \node[below] at (#2,0) {1};
        \node[left] at (0,#3) {1};
        \foreach \x in {1, ..., \value{mx}}
            \node[below] at (\x,0) {\x/#2};
        \foreach \y in {1, ..., \value{my}}
            \node[left] at (0,\y) {\y/#3};
    \end{tikzpicture}
}
\begin{document}
\flattorus{3}{4}
\flattorus{5}{2}
\flattorus[2]{2}{3}
\end{document}

Answer 2

A nice solution was posted by @Vincent (+1) but this was already in the works so I figured I'd post it as an alternative.

Here is a macro \flattorus that takes two arguments (and one additional optional argument) to draw the flat torus. \flattorus[<scale factor>]{y}{x} draws the line of slope y/x on the torus. Default scale is 2 which produces a square of side length 2cm.

For example, \flattorus{3}{4}\qquad\flattorus{5}{3} produces:

And \flattorus[6]{10}{11} produces

\documentclass{article}

\usepackage{tikz}

\newcommand{\flattorus}[3][2]{\begin{tikzpicture}[scale=#1]
    \foreach \k[evaluate=\k as \j using int(\k-1)] in {2,...,#2}{
        \draw[gray!30] ({(\j)/#2},0)node[black, below]{$\frac{\j}{#2}$}--++(0,1);}
    \foreach \k[evaluate=\k as \j using int(\k-1), evaluate=\k as \p using #2*#3] in {2,...,#3}{\xdef\xy{\p}
        \draw[gray!30] (0,{(\j)/#3})node[black, left]{$^{\j}\!/\!_{#3}$}--++(1,0);}
    \foreach \k[evaluate=\k as \j using int(\k-1), evaluate=\k as \x using frac(\k*#3/\xy), evaluate=\k as \y using frac(\k*#2/\xy)] in {1,...,\xy}{
        \draw[red, thick]({frac(\j*#3/\xy)},{frac(\j*#2/\xy)})--({\x+less(\x,1/\xy)},{\y+less(\y,1/\xy)});}
    \draw (0,0)node[below left]{0}--(1,0)node[below]{1}--(1,1)--(0,1)node[left]{1}--cycle;
    \end{tikzpicture}}

\begin{document}

\flattorus{3}{4}\qquad\flattorus{5}{3}

\flattorus[6]{10}{11}

\end{document}

Answer 3

Try this code:

\documentclass[10pt,a4paper]{article}

\usepackage{tikz}
\begin{document}
    \begin{tikzpicture}[scale=10]
        \draw[gray!30,xstep=.3333,ystep=.25] (0,0) grid (1,1);
        
        \draw[red,line width=2pt] (0,0) -- (1,3/4);
        \draw[red,line width=2pt] (0,3/4) -- (1/3,1);
        \draw[red,line width=2pt] (1/3,0) -- (1,1/2);
        \draw[red,line width=2pt] (0,1/2) -- (2/3,1);
        \draw[red,line width=2pt] (2/3,0) -- (1,1/4);
        \draw[red,line width=2pt] (0,1/4) -- (1,1);
        \foreach \x in {0,1/3,2/3,1}
        \draw (\x,.02)--(\x,-.02) node[below] {\bfseries $\x$};
        \foreach \y in {0,1/4,1/2,3/4,1}
        \draw (.02,\y)--(-.02,\y) node[left] {\bfseries $\y$};  
    \end{tikzpicture}
\end{document}

Output: